Hi everyone I'm new to this forum. Im in a class that seems very hard which in my case is Statistics. So been trying to figure out these two problems which I think I start to understand than i get lost. The help would be greatly appreciated(Rofl). 1)Read the clue article. The article mentions all the possible starting hands are not listed. List all possible starting hands. Than calculate how many possible options there are for the correct answers for each of the options. Is it this 4x4x7=112/3x3x6=54/2x2x5=20/1x1x4=4/2x4x6=48/2x3x5=40/1x2x3=6 so is the ones I listed possible starting hands.
2.) professor pigskin is trying to drum up business for his gambling hotline. He is going to give his predictions out free of charge to 16,000 people for three upcoming games this weekend.
A. assuming there are no ties, how many different outcomes are there for the winners of the three games? This one I have no clue unless I take 16C3=560
b. if he sends out all possible predictions evenly, how many of the 16,000 actually receive the correct predictions?? This I don’t know either unless I do the same thing or its just 16,000 This just makes me feel dumb lol

COLONEL MUSTARD WITH THE LEAD PIPE IN THE CONSERVATORY

If you are not familiar with the Parker Brothers board game called CLUE, it is a game where winning involves figuring out which suspect committed the murder, what weapon they used, and in what room the murder took place. In the classic version of the game, there are six suspects (Miss Scarlet, Colonel Mustard, Mrs. White, Mr. Green, Mrs. Peacock, and Professor Plum). There are also six weapons (knife, rope, lead pipe, candlestick, revolver, and wrench) and nine rooms (hall, lounge, dining room, kitchen, ballroom, conservatory, billiard room, library, and study). At the beginning of the game, one suspect, one weapon, and one room are randomly (and secretly) removed from the deck and “hidden” during the game. The game involves determining which three items are “hidden” by trying to find out the rest of the cards in play.
Assuming that there is a full game (six players), each player will be dealt three of the remaining cards. Depending on what cards you are dealt, you may have an easier or harder time winning the game. It all depends on what you are given and how many possible correct answers are left.
If you use all of the cards (6 suspects, 6 weapons, and 9 rooms) and the understanding that you will have one of each in the answer, you multiply these numbers together to determine the number of possible combinations of suspect/weapon/room that could be a possible answer. This would be 6*6*9 = 324 possible options.
Once you know what you have in your hand, you can eliminate some of these options. For example, if you are dealt the “knife” card, you can eliminate all of the 324 options which use the knife as the weapon.
If you are given one suspect, one weapon, and one room card, then that leaves 5 suspects, 5 weapons, and 8 rooms that you do not know and can be used to make the correct answer. This leaves you with 5*5*8 = 200 possible options. Is this a good hand or not? The answer to that can be determined by listing all of the possible sets of cards that you can be dealt to begin a game. Without listing all of the options here, we will just go straight to the “best” starting hand and the “worst” starting hand.
The best hand you could be dealt is either getting three suspect cards (or three weapon cards). That will leave you with three more suspect cards (or three other weapon cards), all six weapon cards (or all six suspect cards), and all nine rooms. This gives you 3*6*9 = 162 possible options….38 fewer than the 200 from the earlier example.
The worst hand you could be dealt is three room cards. This would leave you with all six suspect and weapon cards that you don’t know as well as six of the rooms still to figure out. That means there are still 6*6*6 = 26 options to have to figure out.